Problem: Let $f(x)=7\sin(x)-2x^3$. $f'(x)=$
The expression for $f(x)$ includes $\sin(x)$. Remember that the derivative of $\sin(x)$ is $\cos(x)$. Put another way, $\dfrac{d}{dx}[\sin(x)]=\cos(x)$. $\begin{aligned} f'(x)&=\dfrac{d}{dx}[7\sin(x)-2x^3] \\\\ &=7\dfrac{d}{dx}[\sin(x)]-2\dfrac{d}{dx}(x^3) \\\\ &=7\cdot\cos(x)-2\cdot3x^2 \\\\ &=7\cos(x)-6x^2 \end{aligned}$ In conclusion, $f'(x)=7\cos(x)-6x^2$